Clock math. We say that a\ \text{mod}\ b = r if a=bq+r, such that b>0 and 0 \leq r <b. More specifically, we denote:

if b|(a-a’). additional information basic modular arithmetic operations \begin{align} (a+b)\ \text{mod}\ c &= ((a\ \text{mod}\ c) + (b\ \text{mod}\ c))\ \text{mod}\ c \ (ab) \ \text{mod}\ c &= ((a\ \text{mod}\ c) (b \ \text{mod}\ c)) \ \text{mod}\ c \end{align} examples of modular arithmetic If a\ \text{mod}\ b = r, (-a)\ \text{mod}\ b = -r = b-r 2^{2}\equiv 4 \equiv -1 \ \text{mod}\ 5, 2^{4}\equiv 1\ \text{mod}\ 5 USPS’s check digit is a\ \text{mod}\ 9 because you can just add all the digits up Let a \in \mathbb{Z}. Let s be the sum of all the digits in a. a \ \text{mod}\ 9 = s \ \text{mod}\ 9. Why? Not a very satisfying answer, but because 9 is 10-1, so for each n \times 10^{k}\ \text{mod}\ 9 is always -n smaller. like how 10 = 9+1, 20 = 2 \times 9+2, etc. subgroups Recall the real numbers: \dots, -2, -1, 0, 1, 2, 3, \dots That’s so many numbers! Instead, let’s create a circle of these values. For instance, what if you only want 5:

This is a group under addition. humph: similarity between this and affine subsets u/U = v/U if u-v \in U u \equiv v\ \text{mod}\ b if b|u-v Chinese Remainder Theorem Suppose a,b \in \mathbb{Z}, and m,n \in \mathbb{N}, such that \gcd (m,n) = 1 (that is, suppose m,n is coprime). There is some x \in \mathbb{Z} such that:

Furthermore, and importantly, x\ \text{mod}\ (mn) is unique.